ESTERS

1. REACTIONS OF ORGANIC COMPOUNDS POLYMERS DIBROMOALKANES KETONES ALKANES ALKENES ALCOHOLS ALDEHYDES HALOGENOALKANES ESTERS AMINES NITRILES CARBOXYLIC ACIDS CONVERSIONS

2. REACTIONS OF ORGANIC COMPOUNDS POLYMERS DIBROMOALKANES KETONES P F C S D ALKANES E ALKENES ALCOHOLS M N A G U R T B L V ALDEHYDES HALOGENOALKANES ESTERS O Q U H I T K AMINES NITRILES CARBOXYLIC ACIDS J

3. CHLORINATION OF METHANE A InitiationCl2 ——> 2Cl•radicals created PropagationCl• + CH4 ——> CH3• + HClradicals used and Cl2 + CH3• ——> CH3Cl + Cl•then re-generated TerminationCl• + Cl• ——> Cl2radicals removed Cl• + CH3• ——> CH3Cl CH3• + CH3• ——> C2H6 Summary Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to react Free radicals need to be formed by homolytic fission of covalent bonds This is done by shining UV light on the mixture (heat could be used) Chlorine radicals are produced because the Cl-Cl bond is the weakest You only need one chlorine radical to start things off With excess chlorine you will get further substitution and a mixture of chlorinated products CONVERSIONS

4. ELECTROPHILIC ADDITION OF HBr B Reagent Hydrogen bromide... it is electrophilic as the H is slightly positive Condition Room temperature. EquationC2H4(g) + HBr(g) ———> C2H5Br(l) bromoethane Mechanism Step 1 As the HBr nears the alkene, one of the carbon-carbon bonds breaks The pair of electrons attaches to the slightly positive H end of H-Br. The HBr bond breaks to form a bromide ion. A carbocation (positively charged carbon species) is formed. Step 2 The bromide ion behaves as a nucleophile and attacks the carbocation. Overall there has been addition of HBr across the double bond. CONVERSIONS

5. ELECTROPHILIC ADDITION OF BROMINE C Reagent Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl4 ) Conditions Room temperature. No catalyst or UV light required! EquationC2H4(g) + Br2(l) ——> CH2BrCH2Br(l) 1,2 - dibromoethane Mechanism It is surprising that bromine should act as an electrophile as it is non-polar. CONVERSIONS

6. DIRECT HYDRATION OF ALKENES D Reagent steam Conditions high pressure Catalyst phosphoric acid Product alcohol EquationC2H4(g) + H2O(g) C2H5OH(g) ethanol Use ethanol manufacture Comments It may be surprising that water needs such vigorous conditions to react with ethene. It is a highly polar molecule and you would expect it to be a good electrophile. However, the O-H bonds are very strong so require a great deal of energy to be broken. This necessitates the need for a catalyst. CONVERSIONS

7. HYDROGENATION E Reagent hydrogen Conditions nickel catalyst - finely divided Product alkanes EquationC2H4(g) + H2(g) ———> C2H6(g) ethane Use margarine manufacture CONVERSIONS

8. POLYMERISATION OF ALKENES F EXAMPLES OF ADDITION POLYMERISATION ETHENE POLY(ETHENE) PROPENE POLY(PROPENE) CHLOROETHENE POLY(CHLOROETHENE) POLYVINYLCHLORIDE PVC TETRAFLUOROETHENE POLY(TETRAFLUOROETHENE) PTFE “Teflon” CONVERSIONS

9. NUCLEOPHILIC SUBSTITUTION G AQUEOUSSODIUM HYDROXIDE ReagentAqueous* sodium (or potassium) hydroxide Conditions Reflux in aqueous solution (SOLVENT IS IMPORTANT) Product Alcohol Nucleophile hydroxide ion (OH¯) Equation e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(l) + NaBr(aq) Mechanism *WARNINGIt is important to quote the solvent when answering questions. Elimination takes place when ethanol is the solvent The reaction (and the one with water) is known as HYDROLYSIS CONVERSIONS

10. NUCLEOPHILIC SUBSTITUTION H AMMONIA Reagent Aqueous, alcoholic ammonia (in EXCESS) Conditions Reflux in aqueous, alcoholic solution under pressure Product Amine Nucleophile Ammonia (NH3) Equation e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br (i) C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr (ii) HBr + NH3 (aq / alc) ——> NH4Br Mechanism Notes The equation shows two ammonia molecules. The second one ensures that a salt is not formed. Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE) CONVERSIONS

11. NUCLEOPHILIC SUBSTITUTION H AMMONIA Why excess ammonia? The second ammonia molecule ensures the removal of HBr which would lead to the formation of a salt. A large excess ammonia ensures that further substitution doesn’t take place - see below Problem The amine produced is also a nucleophile (lone pair on N) and can attack another molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and reacts further producing a 3° amine and, eventually a quarternary ammonium salt. C2H5NH2 + C2H5Br ——> HBr + (C2H5)2NH diethylamine, a 2° amine (C2H5)2NH + C2H5Br ——> HBr + (C2H5)3N triethylamine, a 3° amine (C2H5)3N + C2H5Br ——> (C2H5)4N+ Br¯ tetraethylammonium bromide, a 4° salt CONVERSIONS

12. NUCLEOPHILIC SUBSTITUTION I POTASSIUM CYANIDE Reagent Aqueous, alcoholic potassium (or sodium) cyanide Conditions Reflux in aqueous , alcoholic solution Product Nitrile (cyanide) Nucleophile cyanide ion (CN¯) Equation e.g. C2H5Br + KCN (aq/alc) ———> C2H5CN + KBr(aq) Mechanism Importanceit extends the carbon chain by one carbon atom the CN group can then be converted to carboxylic acids or amines. Hydrolysis C2H5CN + 2H2O ———> C2H5COOH + NH3 ReductionC2H5CN + 4[H] ———> C2H5CH2NH2 J K CONVERSIONS

13. ELIMINATION L ReagentAlcoholic sodium (or potassium) hydroxide Conditions Reflux in alcoholic solution Product Alkene Mechanism Elimination EquationC3H7Br + NaOH(alc) ———> C3H6 + H2O + NaBr Mechanism the OH¯ ion acts as a base and picks up a proton the proton comes from a C atom next to the one bonded to the halogen the electron pair moves to form a second bond between the carbon atoms the halogen is displaced; overall there is ELIMINATION of HBr. With unsymmetrical halogenoalkanes, a mixture of products may be formed. CONVERSIONS

14. ELIMINATION OF WATER (DEHYDRATION) L Reagent/catalyst conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4) Conditions reflux at 180°C Product alkene Equation e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l) Mechanism Step 1 protonation of the alcohol using a lone pair on oxygen Step 2 loss of a water molecule to generate a carbocation Step 3 loss of a proton (H+) to give the alkene Note Alcohols with the OH in the middle of a chain can have two ways of losing water. In Step 3 of the mechanism, a proton can be lost from either side of the carbocation. This gives a mixture of alkenes from unsymmetrical alcohols... CONVERSIONS

15. OXIDATION OF PRIMARY ALCOHOLS N Primary alcohols are easily oxidised to aldehydes e.g. CH3CH2OH(l) + [O] ———> CH3CHO(l) + H2O(l) it is essential to distil off the aldehyde before it gets oxidised to the acid CH3CHO(l) + [O] ———> CH3COOH(l) OXIDATION TO ALDEHYDES DISTILLATION OXIDATION TO CARBOXYLIC ACIDS REFLUX Aldehyde has a lower boiling point so distils off before being oxidised further Aldehyde condenses back into the mixture and gets oxidised to the acid CONVERSIONS

16. OXIDATION OF ALDEHYDES O • Aldehydes are easily oxidised to carboxylic acids • e.g. CH3CHO(l) + [O] ———> CH3COOH(l) • one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation • ALDEHYES are EASILYOXIDISED • KETONES are RESISTANTTOMILDOXIDATION • reagents include TOLLENS’ REAGENT and FEHLING’S SOLUTION • TOLLENS’ REAGENT • Reagent ammoniacal silver nitrate solution • Observation a silver mirror is formed on the inside of the test tube • Products silver + carboxylic acid • Equation Ag+ + e- ——> Ag • FEHLING’S SOLUTION • Reagent a solution of a copper(II) complex • Observation a red precipitate forms in the blue solution • Products copper(I) oxide + carboxylic acid • Equation Cu2+ + e- ——> Cu+ CONVERSIONS

17. OXIDATION OF SECONDARY ALCOHOLS P Secondary alcohols are easily oxidised to ketones e.g. CH3CHOHCH3(l) + [O] ———> CH3COCH3(l) + H2O(l) The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment with a powerful oxidising agent they can be further oxidised to a mixture of acids with fewer carbon atoms than the original alcohol. CONVERSIONS

18. REDUCTION OF CARBOXYLIC ACIDS Q Reagent/catalyst lithium tetrahydridoaluminate(III) LiAlH4 Conditions reflux in ethoxyethane Product aldehyde Equation e.g. CH3COOH(l) + 2[H] ———> CH3CHO(l) + H2O(l) CONVERSIONS

19. REDUCTION OF ALDEHYDES R Reagent sodium tetrahydridoborate(III) NaBH4 Conditions warm in water or ethanol Product primary alcohol Equation e.g. C2H5CHO(l) + 2[H]———> C3H7OH(l) CONVERSIONS

20. REDUCTION OF KETONES S Reagent sodium tetrahydridoborate(III) NaBH4 Conditions warm in water or ethanol Product secondary alcohol Equation e.g. CH3COCH3(l) + 2[H]———> CH3CH(OH)CH3(l) CONVERSIONS

21. ESTERIFICATION T Reagent(s) carboxylic acid + strong acid catalyst (e.g conc. H2SO4 ) Conditions reflux Product ester Equation e.g. CH3CH2OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l) NotesConcentrated H2SO4 is also a dehydrating agent, it removes water as it is formed causing the equilibrium to move to the right and thus increasing the yield of ester. Uses of estersEsters are fairly unreactive but that doesn’t make them useless Used as flavourings Naming esters Named from the alcohol and carboxylic acid which made them... CH3OH + CH3COOHCH3COOCH3+ H2O fromethanoic acidCH3COOCH3from methanol METHYLETHANOATE CONVERSIONS

22. HYDROLYSIS OF ESTERS U Reagent(s) dilute acid or dilute alkali Conditions reflux Product carboxylic acid and an alcohol Equation e.g. CH3COOC2H5(l) + H2O(l)CH3CH2OH(l) + CH3COOH(l) Notes If alkali is used for the hydrolysis the salt of the acid is formed CH3COOC2H5(l) + NaOH(aq)———>CH3CH2OH(l) + CH3COO-Na+(aq) CONVERSIONS

23. BROMINATION OF ALCOHOLS V Reagent(s) conc. hydrobromic acid HBr(aq) or sodium (or potassium) bromide and concentrated sulphuric acid Conditions reflux Product haloalkane EquationC2H5OH(l) + conc. HBr(aq) ———> C2H5Br(l) + H2O(l) MechanismThe mechanism starts off in a similar way to dehydration (protonation of the alcohol and loss of water) but the carbocation (carbonium ion) is attacked by a nucleophilic bromide ion in step 3. Step 1 protonation of the alcohol using a lone pair on oxygen Step 2 loss of a water molecule to generate a carbocation (carbonium ion) Step 3 a bromide ion behaves as a nucleophile and attacks the carbocation CONVERSIONS